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I need to show that if $\sum_{k=1}^{\infty}(u_k)^2 \lt 1$ then $\sum_{k=1}^{\infty}\frac{2^{-k}}{1+3^{-k}-u_k} \lt \infty$.

I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.

Can someone maybe help me please?

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  • $\begingroup$ The limit comparison test works: Compare with $2^{-k}$ and use $u_k\to 0.$ $\endgroup$ – zhw. Nov 20 '18 at 17:31
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Note that the convergence of $\sum_{k=1}^{\infty}(u_k)^2$ implies that $u_k\to 0$. Hence $3^{-k}-u_k\to 0$ and there is $N>0$ such that for all $k\geq N$, $|3^{-k}-u_k|<1/2$. Then $$\frac{2^{-k}}{1+1/2}\leq \frac{2^{-k}}{1+3^{-k}-u_k}\leq \frac{2^{-k}}{1-1/2}.$$ Can you take it from here?

P.S. Since $\sum_{k=1}^{\infty}(u_k)^2<1$ then $|u_k|<1$ for each $k\geq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.

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